本试题主要是考查了线面的垂直的证明以及二面角的求解,以及线面平行的判定定理的综合运用
(1)根据已知结合勾股定理和线面垂直的判定定理得到。
(2)建立空间直角坐标系,然后设出点的坐标和向量的坐标,借助于向量的数量积的性质,表示向量的夹角,得到二面角的平面角的求解。
(3)假设存在点PC的中点F, 使得BF//平面AEC.,那个根据假设推理论证,得到结论。
解:(Ⅰ)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248198235.png)
PA =" PD" =" 1" ,PD =" 2" ,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248213195.png)
PA
2 + AD
2 = PD
2, 即:PA ^ AD ---2分
又PA ^ CD , AD , CD 相交于点D,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248213195.png)
PA ^平面ABCD -------4分
(Ⅱ)过E作EG//PA 交AD于G,
从而EG ^平面ABCD,
且AG =" 2GD" , EG = PA = , ------5分
连接BD交AC于O, 过G作GH//OD ,交AC于H,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/201408232232485573584.jpg)
连接EH.
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248198235.png)
GH ^ AC ,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248213195.png)
EH ^ AC ,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248213195.png)
Ð EHG为二面角D—AC―E的平面角. -----6分
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248213195.png)
tanÐEHG = = .
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248213195.png)
二面角D—AC―E的平面角的余弦值为
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248182459.png)
-------7分
(Ⅲ)以AB , AD , PA为x轴、y轴、z轴建立空间直角坐标系.
则A(0 ,0, 0),B(1,0,0) ,C(1,1,0),P(0,0,1),E(0 , ,),
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248822416.png)
= (1,1,0),
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248837423.png)
= (0 , , )
设平面AEC的法向量
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248962310.png)
= (x, y,z) , 则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248978959.png)
,即:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248993857.png)
, 令y =" 1" ,
则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248962310.png)
= (- 1,1, - 2 ) -------------10分
假设侧棱PC上存在一点F, 且
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249040415.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249165400.png)
,
(0 £
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
£ 1), 使得:BF//平面AEC, 则
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249196404.png)
×
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248962310.png)
= 0.
又因为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249196404.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249259407.png)
+
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249040415.png)
= (0 ,1,0)+ (-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
,-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
)= (-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
,1-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
),
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249758187.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249196404.png)
×
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223248962310.png)
=
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
+ 1-
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
- 2
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
=" 0" ,
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249758187.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140823/20140823223249056323.png)
= ,
所以存在PC的中点F, 使得BF//平面AEC. ----------------12分